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Let $A\neq I$ be a $5\times5$ matrix with real entries such that the sum of the entries in each row of $A$ is $1$. Then the sum of all the entries in $A^3$ is

1)$\space 3$ $\qquad $2)$\space 15$ $\qquad$ 3)$\space 5$ $\qquad$ 4)$\space 125$

Solution:

The answer is $5$ if $A = I$, but for $A \neq I$,

I found that vectors $\begin{pmatrix} 1\\ 1\\ 1\\ 1\\ 1\\ \end{pmatrix}$ and $\begin{pmatrix} -1\\ -1\\ -1\\ -1\\ -1\\ \end{pmatrix}$ are eigen vectors for the matrix $A$ corresponding to eigen value $1$. From here, I dont know how to proceed, please help me solve the question..and if there are mistakes in my argument, please tell me..

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Let $e$ be the all one vector.

As you pointed out, we have $Ae=e$

The sum of all the entries in $A^3$ is

$$e^TA^3e=e^TA^2(Ae)=e^TA^2e=e^TAe=e^Te=n$$

where $n$ is equal to 5 for this particular question.

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  • $\begingroup$ Great...explanation...Thank you SiongthyeGoh $\endgroup$ – Sam Christopher May 23 '16 at 6:40
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    $\begingroup$ Really not a fan of using e to describe vectors which aren't a standard unit. A ones vector is easier as 1. $\endgroup$ – Nij May 23 '16 at 7:41

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